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Consider this expression where $A$ and $B$ are matrices, $|i \rangle$ is a ket (column vector) and $\langle j |$ is a bra (row vector) : $$ A | i \rangle \langle j | B \tag1\label1 $$

Due to the general associative properties of the bra-ket notation, this can be interpreted as the inner product of 2 vectors: $$ \left( A | i \rangle \right) \left( \langle j | B \right) \tag2\label2 $$

But by regrouping the terms and considering that outer products can be given a matrix representation, \eqref{1} can also be interpreted as the product of 3 matrices:

$$ A (| i \rangle \langle j |) B \tag3\label3 $$

My confusion comes from the mismatch of the dimensions of expressions \eqref{2} and \eqref{3}. \eqref{2} yields a complex scalar, while \eqref{3} yields a matrix. If the associative property holds, I'd expect the dimensions not to depend on the grouping of the terms. Could somebody please shed some light where I am getting confused?

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It seems you're conflating inner products and outer products. An inner product would be expressed as $$\langle i \vert j \rangle \;\;\; \text{or} \;\;\; \langle i \vert A^\dagger B \vert j\rangle.$$ A column vector times a row vector, $C^{j \times 1} R^{1 \times i}$, results in a $j \times i$ matrix. A row vector times a column vector with compatible dimension, $R^{1 \times j} C^{j \times 1}$, gives a scalar.

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    $\begingroup$ Duh, that was quite silly of me! Thanks! $\endgroup$ – Attila Kun Aug 22 at 0:08

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